Non-homogeneous — method of undetermined coefficients (annihilator method): Exercises

4.6.14 · D4 · Maths › Ordinary Differential Equations › Non-homogeneous — method of undetermined coefficients (annih

Shuru karne se pehle, char symbols ka ek reminder jo hum baar baar use karenge:

Level 1 — Recognition

Goal: ek function dekho aur turant woh operator likho jo use kill karta hai. Abhi solve nahi karna — bas function ko ulta padho aur uska characteristic root nikalo.

L1.1 $g (x) = 7 e^{- 4 x}$ ka annihilator batao.

Recall Solution L1.1

Ek exponential $e^{α x}$ single characteristic root $r = α$ se paida hota hai. Yahan $α = - 4$ hai, toh root $r = - 4$ hai, aur us root wala operator hai $(D - (- 4)) = D + 4$ . Check: $(D + 4) 7 e^{- 4 x} = 7 (- 4) e^{- 4 x} + 4 \cdot 7 e^{- 4 x} = 0. ✓$ Constant $7$ irrelevant hai — annihilators shape ko kill karte hain, size ko nahi. $A = D + 4.$

L1.2 $g (x) = x^{2}$ ka annihilator batao.

Recall Solution L1.2

Powers $1, x, x^{2}, \dots, x^{n}$ sab root $r = 0$ se aate hain jo repeated $n + 1$ baar hota hai, kyunki multiplicity $k$ wala root $0$ produce karta hai $e^{0 x}, x e^{0 x}, \dots = 1, x, \dots, x^{k - 1}$ . $x^{2}$ ke liye $n = 2$ hai, toh $r = 0$ ko $3$ baar repeat karna padega: $A = D^{3}$ . Check: $x^{2}$ ko teen baar differentiate karo: $2 x \to 2 \to 0. ✓$ $A = D^{3} .$

L1.3 $g (x) = sin 5 x$ ka annihilator batao.

Recall Solution L1.3

Frequency $β$ wala pure sine ya cosine pair of complex roots $r = \pm i β$ se aata hai (dekho Characteristic equation and repeated roots). Roots $+ i β$ aur $- i β$ wala polynomial banao: $(r - i β) (r + i β) = r^{2} + β^{2} .$ $β = 5$ ke saath woh hai $A = D^{2} + 25$ . Check: $(D^{2} + 25) sin 5 x = - 25 sin 5 x + 25 sin 5 x = 0. ✓$ $A = D^{2} + 25.$

L1.4 $g (x) = x e^{2 x}$ ka annihilator batao.

Recall Solution L1.4

Extra factor $x$ ka matlab hai root $α = 2$ ko ek aur baar repeat karna padega. Multiplicity $2$ wala root deta hai $e^{2 x}$ aur $x e^{2 x}$ . $x^{n} e^{α x}$ ke liye rule hai multiplicity $n + 1$ ; yahan $n = 1$ hai, toh multiplicity $2$ : $A = (D - 2)^{2} .$ Trap avoid kiya: $(D - 2)$ akele $e^{2 x}$ ko kill karta hai lekin $x e^{2 x}$ ko NAHI — tumhe square chahiye.

Level 2 — Application

Goal: saaf cases par poora six-step algorithm chalao jahan $g$ aur $y_{h}$ ke beech koi overlap NA ho.

L2.1 $y^{''} - y^{'} - 2 y = 6 e^{4 x}$ solve karo.

Recall Solution L2.1

Step 1 — homogeneous. Characteristic equation $r^{2} - r - 2 = (r - 2) (r + 1) = 0$ , roots $2, - 1$ . Toh $y_{h} = c_{1} e^{2 x} + c_{2} e^{- x} .$ Step 2 — annihilate. $g = 6 e^{4 x}$ ka root $4$ hai. Yeh $2$ ya $- 1$ NAHI hai, toh koi overlap nahi. $A = D - 4$ . Step 3–4 — bada homogeneous. Roots ban jaate hain $2, - 1, 4$ . Genuinely naya root $4$ hai, jo deta hai trial $y_{p} = A e^{4 x} .$ Step 5 — determine. $y_{p}^{'} = 4 A e^{4 x}$ , $y_{p}^{''} = 16 A e^{4 x}$ . Substitute karo: $16 A - 4 A - 2 A = 10 A, 10 A e^{4 x} = 6 e^{4 x} \Rightarrow A = \frac{3}{5} .$ Step 6. $y = c_{1} e^{2 x} + c_{2} e^{- x} + \frac{3}{5} e^{4 x} .$

L2.2 $y^{''} + 4 y = 8 x + 3$ solve karo.

Recall Solution L2.2

Step 1. $r^{2} + 4 = 0 \Rightarrow r = \pm 2 i$ , toh $y_{h} = c_{1} cos 2 x + c_{2} sin 2 x$ . Step 2. $g = 8 x + 3$ degree-1 polynomial hai → root $0$ do baar repeat → $A = D^{2}$ . (Koi overlap nahi: $0$ , $\pm 2 i$ nahi hai.) Step 3–4. Naye roots $0, 0$ se terms $1$ aur $x$ milti hain, toh $y_{p} = A x + B .$ Step 5. $y_{p}^{'} = A$ , $y_{p}^{''} = 0$ . Phir $y_{p}^{''} + 4 y_{p} = 4 A x + 4 B = 8 x + 3$ . Match karo: $4 A = 8 \Rightarrow A = 2$ ; $4 B = 3 \Rightarrow B = \frac{3}{4}$ . Step 6. $y = c_{1} cos 2 x + c_{2} sin 2 x + 2 x + \frac{3}{4} .$

L2.3 $y^{''} - 2 y^{'} + 5 y = sin x$ solve karo.

Recall Solution L2.3

Step 1. $r^{2} - 2 r + 5 = 0 \Rightarrow r = \frac{2 \pm 4 - 20}{2} = 1 \pm 2 i$ . Toh $y_{h} = e^{x} (c_{1} cos 2 x + c_{2} sin 2 x) .$ Step 2. $g = sin x$ ke roots $\pm i$ hain (yaani $β = 1$ ). Annihilator $A = D^{2} + 1$ . Koi overlap nahi ( $\pm i \neq = 1 \pm 2 i$ ). Step 3–4. Naye terms $cos x, sin x$ , toh DONO include karna ZAROORI hai (differentiation unhe mix karta hai): $y_{p} = A cos x + B sin x .$ Step 5. $y_{p}^{'} = - A sin x + B cos x$ , $y_{p}^{''} = - A cos x - B sin x$ . $y_{p}^{''} - 2 y_{p}^{'} + 5 y_{p}$ mein substitute karo: $cos x : (- A - 2 B + 5 A) = 4 A - 2 B$ $sin x : (- B + 2 A + 5 B) = 2 A + 4 B$ $sin x$ ke barabar set karo (toh $cos$ -coeff $= 0$ , $sin$ -coeff $= 1$ ): $4 A - 2 B = 0, 2 A + 4 B = 1.$ Pehle se, $B = 2 A$ . Sub karo: $2 A + 8 A = 1 \Rightarrow A = \frac{1}{10}$ , $B = \frac{1}{5}$ . Step 6. $y = e^{x} (c_{1} cos 2 x + c_{2} sin 2 x) + \frac{1}{10} cos x + \frac{1}{5} sin x .$

Level 3 — Analysis (overlap / resonance / sums)

Goal: detect karo jab $g$ ka root $L$ ke saath share hota hai (forcing karta hai extra $x$ ), aur sums ko superposition se handle karo.

L3.1 $y^{''} - 4 y^{'} + 4 y = e^{2 x}$ solve karo.

Recall Solution L3.1

Step 1. $r^{2} - 4 r + 4 = (r - 2)^{2} = 0$ , ek double root $r = 2$ . Toh $y_{h} = c_{1} e^{2 x} + c_{2} x e^{2 x}$ . Step 2 — overlap pakdo. $g = e^{2 x}$ ka root $2$ hai, lekin $2$ pehle se hi $L$ ka root hai multiplicity $2$ ke saath. Annihilator $A = D - 2$ . Apply karo: $(D - 2) (D - 2)^{2} y = (D - 2)^{3} y = 0.$ Step 3–4. Root $2$ ab multiplicity $3$ ka ho gaya, jo deta hai $e^{2 x}, x e^{2 x}, x^{2} e^{2 x}$ . Pehle do exactly $y_{h}$ hain; genuinely naya term hai $x^{2} e^{2 x}$ : $y_{p} = A x^{2} e^{2 x} .$ (Multiplicity 2 se 3 ho gayi, toh naya term $x^{2}$ carry karta hai — isliye hum $x^{m}$ se multiply karte hain jahan $m$ purani multiplicity hai, yahan $m = 2$ .) Step 5. $u = x^{2}$ ke saath: $y_{p} = A x^{2} e^{2 x}$ . $y_{p}^{'} = A e^{2 x} (2 x + 2 x^{2})$ , $y_{p}^{''} = A e^{2 x} (2 + 8 x + 4 x^{2})$ . Phir $y_{p}^{''} - 4 y_{p}^{'} + 4 y_{p} = A e^{2 x} [(2 + 8 x + 4 x^{2}) - 4 (2 x + 2 x^{2}) + 4 x^{2}] = A e^{2 x} \cdot 2.$ Toh $2 A e^{2 x} = e^{2 x} \Rightarrow A = \frac{1}{2}$ . Step 6. $y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{1}{2} x^{2} e^{2 x} .$ Yeh resonance hai exponential bhes mein: system ke apne root par forcing energy pump karti hai, aur response $x^{2}$ ki tarah badhta hai.

L3.2 $y^{''} + 9 y = cos 3 x$ solve karo.

Recall Solution L3.2

Step 1. $r^{2} + 9 = 0 \Rightarrow r = \pm 3 i$ , toh $y_{h} = c_{1} cos 3 x + c_{2} sin 3 x$ . Step 2 — overlap. $g = cos 3 x$ ke roots $\pm 3 i$ hain, wahi jo $L$ ke hain! Annihilator $A = D^{2} + 9$ . Apply karo: $(D^{2} + 9) (D^{2} + 9) y = (D^{2} + 9)^{2} y = 0.$ Step 3–4. Roots $\pm 3 i$ ab multiplicity $2$ ke hain, jo dete hain $cos 3 x, sin 3 x, x cos 3 x, x sin 3 x$ . Pehle do ( $y_{h}$ ) drop karo; naya pair $x$ carry karta hai: $y_{p} = x (A cos 3 x + B sin 3 x) .$ Step 5. $y_{p}^{''}$ compute karo. $y_{p} = A x cos 3 x + B x sin 3 x$ lo. $y_{p}^{''} = A (- 6 sin 3 x - 9 x cos 3 x) + B (6 cos 3 x - 9 x sin 3 x)$ . Phir $y_{p}^{''} + 9 y_{p} = - 6 A sin 3 x + 6 B cos 3 x$ ( $x$ -terms cancel ho jaate hain — yahi resonance ka point hai). $cos 3 x$ se match karo: $6 B = 1 \Rightarrow B = \frac{1}{6}$ ; $- 6 A = 0 \Rightarrow A = 0$ . Step 6. $y = c_{1} cos 3 x + c_{2} sin 3 x + \frac{1}{6} x sin 3 x .$ $x sin 3 x$ term ek badhta hua oscillation hai — textbook resonance.

L3.3 $y^{''} - y = 2 e^{x} + 4$ solve karo (ek sum — superposition use karo).

Recall Solution L3.3

Step 1. $r^{2} - 1 = (r - 1) (r + 1) = 0$ , roots $\pm 1$ . Toh $y_{h} = c_{1} e^{x} + c_{2} e^{- x}$ . Step 2 — superposition se split karo. $L [y] = 2 e^{x}$ aur $L [y] = 4$ alag alag solve karo, phir jodo.

Piece 1: $g_{1} = 2 e^{x}$ , root $1$ — $y_{h}$ ke root $1$ ke saath overlap hai (simple). Toh $x$ se multiply karo: $y_{p 1} = A x e^{x}$ . $y_{p 1}^{'} = A e^{x} (1 + x)$ , $y_{p 1}^{''} = A e^{x} (2 + x)$ . Phir $y_{p 1}^{''} - y_{p 1} = A e^{x} (2 + x) - A x e^{x} = 2 A e^{x}$ . Set karo $2 A e^{x} = 2 e^{x} \Rightarrow A = 1$ . Toh $y_{p 1} = x e^{x}$ .
Piece 2: $g_{2} = 4$ (constant → root $0$ , koi overlap nahi). Trial $y_{p 2} = C$ . Phir $- C = 4 \Rightarrow C = - 4$ . Step 6. $y = c_{1} e^{x} + c_{2} e^{- x} + x e^{x} - 4.$

Level 4 — Synthesis

Goal: sab kuch assemble karo — products ke liye annihilators banao, overlap ko sums ke saath mix karo, aur forms carefully choose karo.

L4.1 Trial form do (sirf unknown coefficients, solve mat karo) for $y^{''} - 2 y^{'} + y = x^{2} e^{x} + sin x,$ given that $L = (D - 1)^{2}$ .

Recall Solution L4.1

$y_{h}$ : $(r - 1)^{2} = 0$ , double root $1$ , toh $y_{h} = c_{1} e^{x} + c_{2} x e^{x}$ . Piece $x^{2} e^{x}$ : annihilator $(D - 1)^{3}$ . Lekin root $1$ pehle se $L$ mein multiplicity $2$ ka hai, toh use apply karne se multiplicity $2 + 3 = 5$ ho jaati hai, jo deta hai $e^{x}, x e^{x}, x^{2} e^{x}, x^{3} e^{x}, x^{4} e^{x}$ . Pehle do ( $y_{h}$ ) drop karo. Naye terms: $x^{2} e^{x}, x^{3} e^{x}, x^{4} e^{x}$ . Trial: $y_{p 1} = (A x^{2} + B x^{3} + C x^{4}) e^{x} .$ (Equivalently: base form $(A + B x + C x^{2}) e^{x}$ times $x^{2}$ double overlap ki wajah se.) Piece $sin x$ : roots $\pm i$ , annihilator $D^{2} + 1$ . Root $1$ ke saath koi overlap nahi. Trial: $y_{p 2} = E cos x + F sin x .$ Combined trial: $y_{p} = (A x^{2} + B x^{3} + C x^{4}) e^{x} + E cos x + F sin x .$ (Chhe unknowns $A, B, C, E, F$ — wait, paanch. Unhe solve karna ek alag exercise hai; yahan hum sirf sahi shape chahte the.)

L4.2 $y^{''} + y = 4 cos x$ (resonant frequency par overlap) solve karo aur interpret karo.

Recall Solution L4.2

Step 1. $r^{2} + 1 = 0 \Rightarrow r = \pm i$ , $y_{h} = c_{1} cos x + c_{2} sin x$ . Step 2 — overlap. $g = 4 cos x$ ke roots $\pm i$ hain, $L$ ke roots ke identical. $A = D^{2} + 1$ , toh $(D^{2} + 1)^{2} y = 0$ , multiplicity 2. Step 3–4. Naye terms: $x cos x, x sin x$ . Toh $y_{p} = x (A cos x + B sin x)$ . Step 5. $y_{p} = A x cos x + B x sin x$ ke saath: $y_{p}^{''} = A (- 2 sin x - x cos x) + B (2 cos x - x sin x)$ . $y_{p}^{''} + y_{p} = - 2 A sin x + 2 B cos x$ . $4 cos x$ se match karo: $2 B = 4 \Rightarrow B = 2$ ; $- 2 A = 0 \Rightarrow A = 0$ . Step 6. $y = c_{1} cos x + c_{2} sin x + 2 x sin x .$ Interpretation. Amplitude $2 x$ bina bound ke badhta hai — pure resonance. Dekho Resonance in forced oscillations. Physically, jhule ko exactly uski natural frequency par push karna use oopar se oopar uthata jaata hai.

L4.3 $y^{'''} - y^{''} = 6$ solve karo (root $0$ par repeated root jo constant forcing ke saath overlap karta hai).

Recall Solution L4.3

Step 1. Characteristic: $r^{3} - r^{2} = r^{2} (r - 1) = 0$ . Roots $0$ (double), $1$ . Toh $y_{h} = c_{1} + c_{2} x + c_{3} e^{x} .$ Step 2 — overlap. $g = 6$ constant hai → root $0$ . Lekin $0$ pehle se $L$ mein multiplicity $2$ ka hai! Annihilator $A = D$ . Apply karo: $D \cdot r^{2} (r - 1) \to r^{3} (r - 1)$ , toh root $0$ ki multiplicity $3$ ho gayi. Step 3–4. Root $0$ ke terms: $1, x, x^{2}$ . Pehle do $y_{h}$ hain; naya wala $x^{2}$ hai. Trial: $y_{p} = A x^{2} .$ Step 5. $y_{p}^{''} = 2 A$ , $y_{p}^{'''} = 0$ . Phir $y_{p}^{'''} - y_{p}^{''} = - 2 A = 6 \Rightarrow A = - 3$ . Step 6. $y = c_{1} + c_{2} x + c_{3} e^{x} - 3 x^{2} .$

Level 5 — Mastery

Goal: method ke baare mein khud reason karo — general statements, degenerate inputs, aur machinery kyun kaam karti hai.

L5.1 Solve kiye bina, sabse chhotey homogeneous equation $A L [y] = 0$ ka order (operator ka degree) batao jo tum banate for $L (D - 1) (D - 2) y = x e^{x} + e^{2 x} .$ Phir batao ki trial $y_{p}$ mein kitne unknown coefficients hain.

Recall Solution L5.1

$L$ ka order $2$ hai (roots $1, 2$ ).

$g_{1} = x e^{x}$ : annihilator $(D - 1)^{2}$ , order $2$ .
$g_{2} = e^{2 x}$ : annihilator $(D - 2)$ , order $1$ . Combined annihilator $A = (D - 1)^{2} (D - 2)$ , order $3$ . Toh $A L$ ka order $2 + 3 = 5$ hai. Trial ke liye overlap accounting:
Root $1$ : $L$ mein originally multiplicity $1$ , forcing ko $(D - 1)^{2}$ chahiye → $A L$ mein multiplicity $1 + 2 = 3$ . Terms $e^{x}, x e^{x}, x^{2} e^{x}$ ; $e^{x}$ ( $y_{h}$ ) drop karo, naye: $x e^{x}, x^{2} e^{x}$ → base $(A + B x) e^{x} \cdot x$ se coefficients.
Root $2$ : originally multiplicity $1$ , forcing $(D - 2)$ → multiplicity $2$ . Terms $e^{2 x}, x e^{2 x}$ ; $e^{2 x}$ drop karo, naya: $x e^{2 x}$ . Toh $y_{p} = (A x + B x^{2}) e^{x} + C x e^{2 x}$ : 3 unknown coefficients $A, B, C$ .

L5.2 Degenerate input. $g (x) = 0$ ka annihilator kya hai? Method phir kya produce karta hai, aur kya yeh consistent hai?

Recall Solution L5.2

Zero function ko identity operator... trivially kuch bhi annihilate karta hai, aur uska lowest-order annihilator woh case hai jahan "zero tak pahunchne ke liye kuch nahi karna": literally $A = 1$ (identity times a constant), kyunki $1 \cdot 0 = 0$ aur koi lower order exist nahi karta. $A = 1$ ko $L [y] = 0$ par apply karne se $L [y] = 0$ unchanged milta hai — koi naye roots nahi. Toh $y_{p}$ mein koi naye terms nahi: $y_{p} = 0$ . Consistency: equation $L [y] = 0$ IS homogeneous hai; uska poora solution sirf $y_{h}$ hai. Method correctly $y = y_{h} + 0$ return karta hai. Sab agree karta hai. ✓

L5.3 Mechanism explain karo. Ek paragraph mein argue karo ki $L$ ke roots se aane wale terms (bade equation $A L [y] = 0$ ke andar) $y_{p}$ banane mein kabhi help kyun nahi kar sakte.

Recall Solution L5.3

Koi bhi term $ϕ$ jo $L$ ke root se generate hoti hai, by definition $L [ϕ] = 0$ satisfy karti hai — yahi "root of $L$ " ka matlab hai (dekho Characteristic equation and repeated roots). Toh agar hum aisi $ϕ$ ko $y_{p}$ ke andar use karne ki koshish karein, use original equation mein plug karne par $L [ϕ] = 0$ milta hai, jo required right-hand side $g$ mein kuch contribute nahi karta. Sirf naye roots jo annihilator $A$ add karta hai woh terms $ψ$ produce karte hain jahan $L [ψ] \neq = 0$ , aur wahi $g$ ko match karne capable hain. Isliye $y_{p}$ purely " $A$ -part" se banta hai, aur " $L$ -part" discard ho jaata hai kyunki woh sirf $y_{h}$ reconstruct karta hai. Yahi wajah hai ki overlap $x$ ko force karta hai: shared root $A L$ mein multiplicity gain karta hai, aur extra multiplicity ka term higher power of $x$ carry karta hai jo genuinely naya hai (pehle se $y_{h}$ mein nahi).

L5.4 Initial conditions $y (0) = 0, y^{'} (0) = 1$ ke saath completely solve karo: $y^{''} - 3 y^{'} + 2 y = 4 e^{3 x} .$ (Yeh parent note ka Worked Example 1 hai, ab IVP ke saath — mastery step constants pin karna hai.)

Recall Solution L5.4

Parent note se, general solution hai $y = c_{1} e^{x} + c_{2} e^{2 x} + 2 e^{3 x} .$ $y (0) = 0$ apply karo: $c_{1} + c_{2} + 2 = 0$ . $y^{'} = c_{1} e^{x} + 2 c_{2} e^{2 x} + 6 e^{3 x}$ ; $y^{'} (0) = 1$ apply karo: $c_{1} + 2 c_{2} + 6 = 1$ . Doosri equation mein se pehli ghataao: $c_{2} + 4 = 1 \Rightarrow c_{2} = - 3$ . Phir $c_{1} = - 2 - c_{2} = - 2 + 3 = 1$ . $y = e^{x} - 3 e^{2 x} + 2 e^{3 x} .$ Verify: $y (0) = 1 - 3 + 2 = 0$ ✓; $y^{'} (0) = 1 - 6 + 6 = 1$ ✓.

Wrap-up

Recall Poore page ka one-line summary

$g$ ko ulta padho uske roots tak → $A$ likho → $A L [y] = 0$ banao → sirf woh roots rakho jo $A$ ne add kiye (shared root ki multiplicity $m$ ke liye $x^{m}$ se multiply karo) → real equation mein coefficients solve karo → $y_{h}$ jodo → phir (agar diya ho) total par initial conditions apply karo.